Consider a solid rod of mass $m$ and uniform density resting against a vertical wall and horizontal ...

For equilibrium, the sum of vertical and horizontal forces must be zero, and the sum of torques about any point must also be zero.$$ \begin{aligned} &\text { From the diagram, we can see that: }\\ &N_2+f_1=m g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N_1=f_2=\mu_2 N_2\end{aligned} $$$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \begin{aligned} & f_1=\mu_1 N_1=\mu_1 \mu_2 N_2 \\ & N_2\left(1+\mu_1 \mu_2\right)=m g \\ & \Rightarrow N_2=\frac{m g}{\left(1+\mu_1 \mu_2\right)} \end{aligned} $$Torque Balance:Balancing the torque, we get,$$ f_1 l \cos (\alpha)+N_1 l \sin (\alpha)=m g \frac{l}{2} \cos (\alpha) $$where $l$ is the length of the rod.$$ \begin{gathered} f_1+N_1 \tan (\alpha)=\frac{m g}{2} \\ N_1 \tan (\alpha)=\frac{m g}{2}-f_1=\frac{m g}{2}-\mu_1 N_1 \end{gathered} $$$$ \begin{aligned} &\Rightarrow \tan (\alpha)=\left(\frac{m g}{2 N_1}-\mu_1\right)\\ &\text { So, the angle of inclination } \alpha \text {, such that the rod stays in equilibrium is }\\ &\begin{gathered} \alpha=\tan ^{-1}\left(\frac{1+\mu_1 \mu_2}{2 \mu_2}-\mu_1\right) \\ \alpha=\tan ^{-1}\left(\frac{1-\mu_1 \mu_2}{2 \mu_2}\right) \end{gathered} \end{aligned} $$

Rotational Motion - IAT-IISER Physics PYQs

Q: An inextensible cord of negligible mass passes over the rim of a solid disc of mass $M$ and radius $...

For the system described, two blocks of masses $M$ and $M/2$ are connected by an inextensible cord over a solid disc, and we're tasked with finding the angular acceleration of the disc. Let's go through the process:Tensions in the cord:For mass $M$:$ M g - T_1 = M a \quad \Rightarrow \quad T_1 = M(g - a) \quad \text{(Equation 1)} $For mass $M/2$:$ T_2 - \frac{M}{2} g = \frac{M}{2} a \quad \Rightarrow \quad T_2 = \frac{M}{2}(g - a) \quad \text{(Equation 2)} $Relation between linear and angular accelerations:Since the cord is inextensible and there's no slipping, the linear acceleration $a$ of the blocks relates to the angular acceleration $\alpha$ of the disc by:$ a = R \alpha $Torque on the disc:The torque due to the tensions on the disc is given by:$ \text{Torque} = T_1 R - T_2 R = I \alpha $The moment of inertia $I$ for the disc rotating about its center is:$ I = \frac{1}{2} M R^2 $Thus,:$ (T_1 - T_2) R = \frac{1}{2} M R^2 \alpha \quad \text{(Equation 3)} $Substitute tensions into the torque equation:Substituting Equations 1 and 2 into Equation 3, we have:$ M(g - R \alpha) - \frac{M}{2}(R \alpha + g) = \frac{1}{2} M R \alpha $Simplify and solve for $\alpha$:Simplify the equation:$ M g - M R \alpha - \frac{M}{2} R \alpha - \frac{M}{2} g = \frac{1}{2} M R \alpha $Arrange and combine terms:$ 2g - 2R \alpha - (R \alpha + g) = R \alpha $Simplify further:$ 2g - 2R \alpha - R \alpha - g = R \alpha $$ g = 4R \alpha $Thus, the angular acceleration $\alpha$ is:$ \alpha = \frac{g}{4R} $

For equilibrium, the sum of vertical and horizontal forces must be zero, and the sum of torques about any point must also be zero.$

%5Cbegin%7Baligned%7D%20%26%5Ctext%20%7B%20From%20the%20diagram%2C%20we%20can%20see%20that%3A%20%7D%5C%5C%20%26N_2%2Bf_1%3Dm%20g%20%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2CN_1%3Df_2%3D%5Cmu_2%20N_2%5Cend%7Baligned%7D

%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%20%5Cbegin%7Baligned%7D%20%26%20f_1%3D%5Cmu_1%20N_1%3D%5Cmu_1%20%5Cmu_2%20N_2%20%5C%5C%20%26%20N_2%5Cleft(1%2B%5Cmu_1%20%5Cmu_2%5Cright)%3Dm%20g%20%5C%5C%20%26%20%5CRightarrow%20N_2%3D%5Cfrac%7Bm%20g%7D%7B%5Cleft(1%2B%5Cmu_1%20%5Cmu_2%5Cright)%7D%20%5Cend%7Baligned%7D

Torque%20Balance%3ABalancing%20the%20torque%2C%20we%20get%2C

f_1%20l%20%5Ccos%20(%5Calpha)%2BN_1%20l%20%5Csin%20(%5Calpha)%3Dm%20g%20%5Cfrac%7Bl%7D%7B2%7D%20%5Ccos%20(%5Calpha)

where

is%20the%20length%20of%20the%20rod.

%5Cbegin%7Bgathered%7D%20f_1%2BN_1%20%5Ctan%20(%5Calpha)%3D%5Cfrac%7Bm%20g%7D%7B2%7D%20%5C%5C%20N_1%20%5Ctan%20(%5Calpha)%3D%5Cfrac%7Bm%20g%7D%7B2%7D-f_1%3D%5Cfrac%7Bm%20g%7D%7B2%7D-%5Cmu_1%20N_1%20%5Cend%7Bgathered%7D

%5Cbegin%7Baligned%7D%20%26%5CRightarrow%20%5Ctan%20(%5Calpha)%3D%5Cleft(%5Cfrac%7Bm%20g%7D%7B2%20N_1%7D-%5Cmu_1%5Cright)%5C%5C%20%26%5Ctext%20%7B%20So%2C%20the%20angle%20of%20inclination%20%7D%20%5Calpha%20%5Ctext%20%7B%2C%20such%20that%20the%20rod%20stays%20in%20equilibrium%20is%20%7D%5C%5C%20%26%5Cbegin%7Bgathered%7D%20%5Calpha%3D%5Ctan%20%5E%7B-1%7D%5Cleft(%5Cfrac%7B1%2B%5Cmu_1%20%5Cmu_2%7D%7B2%20%5Cmu_2%7D-%5Cmu_1%5Cright)%20%5C%5C%20%5Calpha%3D%5Ctan%20%5E%7B-1%7D%5Cleft(%5Cfrac%7B1-%5Cmu_1%20%5Cmu_2%7D%7B2%20%5Cmu_2%7D%5Cright)%20%5Cend%7Bgathered%7D%20%5Cend%7Baligned%7D

For the system described, two blocks of masses

M

and

M%2F2

are connected by an inextensible cord over a solid disc, and we're tasked with finding the angular acceleration of the disc. Let's go through the process:Tensions in the cord:For mass

M

M%20g%20-%20T_1%20%3D%20M%20a%20%5Cquad%20%5CRightarrow%20%5Cquad%20T_1%20%3D%20M(g%20-%20a)%20%5Cquad%20%5Ctext%7B(Equation%201)%7D

For mass

M%2F2

T_2%20-%20%5Cfrac%7BM%7D%7B2%7D%20g%20%3D%20%5Cfrac%7BM%7D%7B2%7D%20a%20%5Cquad%20%5CRightarrow%20%5Cquad%20T_2%20%3D%20%5Cfrac%7BM%7D%7B2%7D(g%20-%20a)%20%5Cquad%20%5Ctext%7B(Equation%202)%7D

Relation between linear and angular accelerations:Since the cord is inextensible and there's no slipping, the linear acceleration

a

of the blocks relates to the angular acceleration

%5Calpha

of the disc by:

a%20%3D%20R%20%5Calpha

Torque on the disc:The torque due to the tensions on the disc is given by:

%5Ctext%7BTorque%7D%20%3D%20T_1%20R%20-%20T_2%20R%20%3D%20I%20%5Calpha

The moment of inertia

I

for the disc rotating about its center is:

I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%5E2

Thus,:

(T_1%20-%20T_2)%20R%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%5E2%20%5Calpha%20%5Cquad%20%5Ctext%7B(Equation%203)%7D

Substitute tensions into the torque equation:Substituting Equations 1 and 2 into Equation 3, we have:

M(g%20-%20R%20%5Calpha)%20-%20%5Cfrac%7BM%7D%7B2%7D(R%20%5Calpha%20%2B%20g)%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%20%5Calpha

Simplify and solve for

%5Calpha

:Simplify the equation:

M%20g%20-%20M%20R%20%5Calpha%20-%20%5Cfrac%7BM%7D%7B2%7D%20R%20%5Calpha%20-%20%5Cfrac%7BM%7D%7B2%7D%20g%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20M%20R%20%5Calpha

Arrange and combine terms:

2g%20-%202R%20%5Calpha%20-%20(R%20%5Calpha%20%2B%20g)%20%3D%20R%20%5Calpha

Simplify further:

2g%20-%202R%20%5Calpha%20-%20R%20%5Calpha%20-%20g%20%3D%20R%20%5Calpha

g%20%3D%204R%20%5Calpha

Thus, the angular acceleration

%5Calpha

is:

%5Calpha%20%3D%20%5Cfrac%7Bg%7D%7B4R%7D

Rotational Motion

High-Yield Trend

Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Rotational Motion

High-Yield Trend

Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Questions
3 MCQs